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Suppose I am solving the following limit:

$$\lim_{(x,y)\rightarrow(1, -1)} \frac{x^2 - y^2}{x^3 + y^3}$$

I begin by factorizing as follows:

$$\lim_{(x,y)\rightarrow(1,-1)} \frac{(x+y)(x-y)}{(x+y)(x^2-xy+y^2)}$$

And then I am tempted to factor out $(x+y)$ from the numerator and denominator:

$$\lim_{(x,y)\rightarrow(1,-1)} \frac{(x-y)}{(x^2-xy+y^2)}$$

Normally when factoring out the terms we must check that they do not equal zero. Sometimes in the context of limits though, we can disregard this concern if the term being factored out is only approaching zero without ever reaching it. However in the example above it seems that we must still consider this, since if we approach $(1,-1)$ along the line $x = -y$, then the term $(x + y)$ does indeed equal 0, even before it reaches the origin.

What is going on here?

1) When evaluating the limit in $\mathbb{R}^2$, do we only care about approach paths which are actually defined as the approach the point in question? (the path $(x = -y)$ in this case is undefined at all points!)

2) Does the existence of this approach path for which a limit doesn't exist (because nothing on that path is defined anyway!) mean that the limit itself at that point doesn't exist?

Brendan Hill
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1 Answers1

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IMO unless you state $x \ne -y$, this limit does not exist. However, if you state $x \ne -y$, then limit exists and equals to $\frac{2}{3}$

crbah
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  • In this case I think that "stating" the obvious is of not help: if $;x=-y;$ then there is no mathematical expression to consider to begin with, so I'm supposing the expression already requires $;x\neq -y;$ . – DonAntonio Mar 17 '16 at 12:00
  • @Joanpemo you can discuss what it means but "undefined" is a mathematical terminology. – crbah Mar 17 '16 at 12:09
  • $x \ne y$ is not stated in the problem as the definition of $f(x,y)$ so this restriction has not been made explicit. – Brendan Hill Mar 17 '16 at 12:10
  • BTW wolfram alpha says that the limit does exist and equals $\frac{2}{3}$ - is it unreliable in these cases? http://www.wolframalpha.com/input/?asynchronous=false&equal=Submit&i=limit+as+(x,y)+-%3E+(1,+-1)+of+(x%5E2-y%5E2)%2F(x%5E3%2By%5E3) – Brendan Hill Mar 17 '16 at 12:10
  • @BrendanHill that means wolfram alpha already assumes $x \ne -y$ – crbah Mar 17 '16 at 12:11
  • There are usefull links on discussions on what mathematically means 'undefined'. http://math.stackexchange.com/questions/1228586/what-does-the-term-undefined-actually-mean, http://mathandmultimedia.com/2010/04/21/the-definition-of-undefined/, https://en.wikipedia.org/wiki/Undefined_(mathematics) – crbah Mar 17 '16 at 12:13
  • I think WA is taking as obvious the fact that if you write $;1/x;$ then you already mean beforehand that $;x\neq0$ . This kind of things are hammered into kids in highs school's (at least at some over here), yet hardly someone at a higher level would assume that something funny or misterious happens when $;x=0$ , say. In the present case, there is no mathematical expression (within the framewokr of mathematics I am aware of) for $;\frac{x^2-y^2}{x^3+y^3};$ whenever $;x=-y;$ , In the cases I know of, we always assume the limit's taken within the natural, or given, def. domain. – DonAntonio Mar 18 '16 at 14:01